SUMS OF SEVENTH POWERS IN THE RING
OF POLYNOMIALS OVER THE FINITE FIELD
WITH FOUR ELEMENTS
MIREILLE CAR
Abstract. We study representations of polynomials P ∈ F4 [T ] as sums P = X17 + . . . + Xs7 .
1. Introduction
Let F be a finite field of characteristic p with q = pm elements. Analogues of the Waring’s problem
for the polynomial ring F [T ] were investigated, ( [19], [12], [16], [6], [17], [8], [5], [13], [14], [10],
[9], [2], [3] [4]). Let k > 1 be an integer. Roughly speaking, Waring’s problem over F [T ] consists
in representing a polynomial M ∈ F [T ] as a sum
(1.1)
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M = M1k + . . . + Msk
with M1 , . . . , Ms ∈ F [T ]. Some obstructions to that may occur ([15]), and lead to consider
Waring’s problem over the subring S(F [T ], k) formed by the polynomials of F [T ] which are sums
of k-th powers. Some cancellations may occur in representations (1.1), so that it is possible to
have a representation (1.1) with deg M small and deg(Mik ) large. Without degree conditions in
(1.1), the problem of representing M as sum (1.1) is close to the so called easy Waring’s problem
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Received March 6, 2012.
2010 Mathematics Subject Classification. Primary 11T55; Secondary 11P05.
Key words and phrases. Waring’s problem; polynomials.
for Z. In order to have a problem close to the non-easy Waring’s problem, the degree conditions
(1.2)
k deg Mi < deg M + k
are required. Representations (1.1) satisfying degree conditions (1.2) are called strict representations, see [6, Definition 1.8] in opposition to representations without degree conditions. For the
strict Waring’s problem, analogue of the classical Waring numbers gN (k) and GN (k) have been
defined as follows. Let g(pm , k) denote the least integer s (if it exists) such that every polynomial
M ∈ S(F [T ], k) may be written as a sum (1.1) satisfying the degree conditions (1.2); otherwise we
put g(pm , k) = ∞.
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Similarly, G(pm , k) denotes the least integer fulfilling the above condition for each polynomial
M ∈ S(F [T ], k) of sufficiently large degree. This notation is possible since these numbers depend
only on pm and k. The set S(F [T ], k) and the parameters G(pm , k), g(pm , k) are not sufficient
to describle all possible cases, see [1, Proposition 4.4], so that in [2] and [3] we introduced new
parameters defined as follows.
Let S × (F [T ], k) denote the set of polynomials in F [T ] which are strict sums of k-th powers.
Let g × (pm , k) denote the least integer s (if it exists) such that every polynomial M ∈ S × (F [T ], k)
may be written as a strict sum
M = M1k + . . . + Msk .
Similarly, G× (pm , k) denotes the least integer s fulfilling the same condition for each polynomial
M ∈ S × (F [T ], k) of sufficiently large degree. Gallardo’s method for cubes ([8] and [5]) was
generalized in [1] and [11] where bounds for g(pm , k) and G(pm , k) were established when pm and
k satisfy some conditions. A bound for g(pm , k) was established in [1] in the case when F = S(F, k)
if one of the two following conditions is satisfied:
i) p > k
ii) pn > k = hpν − 1 for some integers ν > 0 and 0 < h ≤ p.
The smallest exponent k satisfying condition ii) is k = 3. It gave a matter for many articles,
see [8], [5], [9], [10]. In the case of even characteristic, the second smallest exponent k satisfying
condition ii) is k = 7. The case k = 7, q = 2m with m > 3 is covered by [1, Theorems 1.2 and 1.3]
or by [11, Theorem 1.4]. For almost all q = 2m , the upper bounds obtained in these articles for
the numbers G(2m , 7) are comparable with the bound GN (7) ≤ 33 known for the corresponding
Waring’s number for the integers ([18]). The case of the numbers g(2m , 7) is different. In the
case when m ∈
/ {1, 2, 3} [1, Theorem 1.3] as well as [11, Theorem 1.4] gives g(2m , 7) ≤ 239`(2m , 7)
when for the integers, it is known that gN (7) = 143 ([7]). In [4] we obtained better bounds for the
numbers g(2m , 7) in the case when m ∈
/ {1, 2, 3}, the method yielding also to better bounds for
some numbers G(2m , 7). The aim of this paper is the study of one of the remaining cases, namely,
the case q = 4. The case q = 8 will be the subject of a separate paper. When a finite field with
8 elements is not a 7-Waring field, every field with 4h elements is a 7-Waring field, so that, from
[15], S(F4 [T ], 7) = F4 [T ]. We will see further that T is not a strict sum of seventh powers in the
ring F4 [T ], see Proposition 3.5 below, so that S(F4 [T ], 7) 6= S × (F4 [T ], 7).
The main results proved in this work are summarized in the following theorems.
Theorem 1.1. We have
S × (F4 [T ], 7) = A1 ∪ A2 ∪ A3 ∪ A∞ ,
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where
(i) A1 is the set of polynomials A =
7
P
n=0
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an T n ∈ F4 [T ] such that a1 = a4 , a2 = a5 , a3 = a6 ;
(ii) A2 is the set of polynomials A =
14
P
an T n ∈ F4 [T ] with 7 < deg A ≤ 14 such that
n=0

 a1 + a4 + a10 + a13 = 0,
a2 + a5 + a8 + a11 = 0,

a3 + a6 + a9 + a12 = 0;
(iii) A3 is the set of polynomials A =
21
P
an T n ∈ F4 [T ] with 14 < deg A ≤ 21 such that
n=0
a3 + a6 + a9 + a12 + a15 + a18 = 0;
(iv) A∞ = {A ∈ F4 [T ] | deg A > 21}.
See Proposition 6.6 below.
Theorem 1.2. Every polynomial P ∈ F4 [T ] with degree ≥ 435 is a strict sum of 33 seventh
powers, so that
G(4, 7) = G× (4, 7) ≤ 33,
and we have
g(4, 7) = ∞,
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×
g (4, 7) ≤ 43.
This theorem is given by Corollaries 3.6, 6.4 and by Theorem 6.7
Proving that polynomials of small degree are sums or strict sums of seventh powers requires
some results on the solvability of systems of algebraic equations over the finite field F4 . This is
done in Section 2. A characterization of polynomials of degree ≤ 21 that are strict sums of seventh
powers is given in Section 3. In Section 4, using the general descent process described in [1], we
obtain a first upper bound for G(4, 7). In Section 5 we describe other descent processes. They are
used in Section 6 to get a better upper bound for G(4, 7) as well as a bound for g(4, 7). We denote
by F the field F4 and by α a root of the equation α2 = α + 1 .
2. Equations
Proposition 2.1. For every (a, b) ∈ F 2 , the system
x1 + x2 = a,
(A(a, b))
u1 x1 + u2 x2 = b,
has solutions (u1 , u2 , x1 , x2 ) ∈ F 4 satisfying the condition x1 x2 u1 u2 6= 0.
Proof. Suppose a = b. Choose x1 ∈ F − {0, a}. Then, (1, 1, x1 , a +x1 ) is a solution of (A(a,
b)).
au2 +b
au2 +b
Suppose a 6= b. There is u2 ∈ F − F2 such that au2 + b 6= 0. Then, 1, u2 , 1+u2 , a + 1+u2 is a
2 +b
solution of (A(a, b)). Moreover, since a 6= b, we have au
1+u2 6= a, so that
au2 + b
au2 + b
× a+
6= 0.
u2 ×
1 + u2
1 + u2
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Proposition 2.2. For (a, b, c) ∈ F 3 , let (Bs (a, b, c)) denote the system of equations

 x1 + . . . + xs = a,
y1 + . . . + ys = b,

x1 y1 + . . . + xs ys = c.
(I) For every (a, b, c) ∈ F × ×F ×F , the system (B2 (a, b, c)) admits solutions (x1 , x2 , y1 , y2 ) ∈ F 4
satisfying the condition x1 x2 6= 0.
(II) For every (a, b, c) ∈ F 3 , the system (B3 (a, b, c)) admits solutions (x1 , x2 , x3 , y1 , y2 , y3 ) ∈ F 6
satisfying the condition x1 x2 x3 y1 y2 y3 6= 0.
(III) For every (a, b, c) ∈ F × × F × F , the system (B3 (a, b, c)) admits solutions (x1 , x2 , x3 , y1 ,
y2 , y3 ) ∈ F 6 satisfying the conditions
x1 x2 x3 y1 y2 y3 6= 0,
x21 y1 6= x22 y2 .
Proof. (I) Suppose a 6= 0. Let x1 ∈ F − {0, a} and let x2 = a + x1 . Then, x2 6= 0 and x2 6= x1 .
The matrix
1
1
x1 x2
is invertible. Thus, for each (b, c) ∈ F 2 , there exists (y1 , y2 ) ∈ F 2 such that
y1 + y2 = b,
x1 y1 + x2 y2 = c.
(II) Let E(a, b, c) denote the set of (x1 , x2 , x3 , y1 , y2 , y3 ) ∈ F 6 solutions of (B3 (a, b, c)) satisfying
x1 x2 x3 y1 y2 y3 6= 0, and satisfying
x1 x2 x3 y1 y2 y3 6= 0,
x21 y1 6= x22 y2 ,
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respectively. For (x1 , x2 , x3 , y1 , y2 , y3 ) ∈ F 6 , the three following statements are equivalent:
(i) (x1 , x2 , x3 , y1 , y2 , y3 ) ∈ E(a, b, c),
(ii) (y1 , y2 , y3 , x1 , x2 , x3 ) ∈ E(b, a, c),
(iii) (x1 y1 , x2 y2 , x3 y3 , (y1 )2 , (y2 )2 , (y3 )2 ) ∈ E(c, b2 , a). Thus, it suffices to deal with the cases
(a, b, c) = (0, 0, 0), (a, b, c) = (a, 0, 0) with a 6= 0, (a, b, c) = (a, b, 0) with ab 6= 0, and
(a, b, c) with abc 6= 0. Firstly, we observe that if x ∈ F − F2 , then (1, x, x + 1, 1, x, x + 1) ∈
E(0, 0, 0). Now, we consider the systems with a 6= 0. Up to the automorphism x 7→ ax,
and the F2 -automorphism α 7→ α + 1, it suffices to consider the cases (a, b, c) = (1, 0, 0),
(a, b, c) = (1, 1, 0), (a, b, c) = (1, 1, 1), (a, b, c) = (1, 1, α). Observe that
(1, 1, 1, 1, α, α + 1) ∈ E(1, 0, 0),
(α + 1, α + 1, 1, 1, α, α) ∈ E(1, 1, 0),
(1, α, α, 1, 1, 1) ∈ E(1, 1, 1),
(1, α + 1, α + 1, α + 1, α + 1, 1) ∈ E(1, 1, α).
Proposition 2.3. For every (a, b, c) ∈ F 3 , the system

 x1 + x2 = a,
y1 + y2 = b,
(C(a, b, c))

x1 y1 z12 + x21 y12 + x2 y2 z22 + x22 y22 = c
admits solutions (x1 , x2 , y1 , y2 , z1 , z2 ) ∈ F 6 satisfying the condition x1 x2 y1 y2 6= 0.
Proof. Let x1 ∈ F be such that x1 6= 0, a, let y1 ∈ F be such that y1 6= 0, b and let z1 ∈ F .
Let x2 = a + x1 and y2 = b + y1 . Then, x1 x2 y1 y2 6= 0. Let z2 ∈ F be defined by the relation
x22 y22 z2 = c2 + x21 y12 z1 + x1 y1 + x2 y2 . Then, (x1 , x2 , y1 , y2 , z1 , z2 ) is a solution of (C(a, b, c)).
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Lemma 2.4. For every (a, b, c) ∈ F × × F × F , the system of equations
z1 + αz2 + (α + 1)z3 = b,
(S1 (a, b, c))
az1 + z12 + (α + 1)az2 + z22 + αaz3 + z32 = c,
admits solutions (z1 , z2 , z3 ) ∈ F 3 .
Proof. Let ν = ν(a, b, c) denote the number of (z1 , z2 , z3 ) ∈ F 3 solutions of (S1 (a, b, c)). For
t ∈ F let
Ψ(t) = (−1)tr(t)
where tr : F → F2 is the absolute trace map. Then Ψ is a non-trivial character, so that by
orthogonality,
X
1X
ν=
Ψ(t(b + z1 + αz2 + (α + 1)z3 ))
4
3
t∈F
(z1 ,z2 ,z3 )∈F
1X
×
Ψ(u(c + az1 + z12 + (α + 1)az2 + z22 + αaz3 + z32 )).
4
u∈F
Thus,
!
X
16ν =
Ψ(bt + cu)
(t,u)∈F 2
X
2
Ψ((t + au)z + uz )
z∈F
!
×
X
2
Ψ((αt + a(α + 1)u)z + uz )
z∈F
!
×
X
2
Ψ(((α + 1)t + αau)z + uz ) .
z∈F
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From [2, Proposition 2.3], for (v, w) ∈ F 2 , we have
X
4
Ψ(vz + wz 2 ) =
0
z∈F
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if
if
w = v2 ,
w 6= v 2 .
Therefore,
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ν=4
X
(t,u)∈E
Ψ(bt + cu),
where E is the subset of F 2 formed by the pairs (t, u) satisfying the three conditions

 u = (t + au)2 ,
u = (αt + a(α + 1)u)2 ,

u = ((α + 1)t + αau)2 .
Obviously, (0, 0) ∈ E. Conversely, let (t, u) ∈ E. Then the first and second conditions give that
t + au = αt + a(α + 1)u while the first and last conditions give that t + au = (α + 1)t + αau, so
that (α + 1)au = t = αau with a 6= 0. Thus, t = u = 0, so that E = {(0, 0)} and ν = 4.
Proposition 2.5. Let b = (a, b, c, d) ∈ F 4 . Then the system of equations
 3
P


yi = a,



i=1


3
P


ui yi = b,

i=1
(D(b))
3
P



u2i zi2 yi = c,


i=1


P
3


 (u2i zi + ui zi2 )yi = d
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i=1
admits solutions (u1 , u2 , u3 , y1 , y2 , y3 , z1 , z2 , z3 ) ∈ F 9 such that u1 u2 u3 y1 y2 y3 6= 0.
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Proof. (I) Suppose that there exists (y1 , y2 , y3 , u) ∈ F 4 satisfying the conditions:

y1 + y2 + y3 = a,



y1 + y2 + uy3 = b,
y1 y2 y3 u 6= 0,



y1 6= y2 ,
and denote (H) this hypothesis. Then the matrix
y1
y12
y2
y22
is invertible. Let z3 ∈ F . There is (z1 , z2 ) ∈ F 2 such that
y1 z1 + y2 z2 = c + d + (u2 z3 + u2 z32 + uz32 )y3 ,
y12 z1 + y22 z2 = c2 + uz3 y32 .
Then, we have
z12 y1 + z22 y2 + u2 z32 y3 = c,
(z1 + z12 )y1 + (z2 + z22 )y2 + (u2 z3 + uz32 )y3 = d,
so that (1, 1, u, y1 , y2 , y3 , z1 , z2 , z3 ) is a solution of (D(b)) such that uy1 y2 y3 6= 0.
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(II) We prove that if one of the three following conditions:
(i) a = b,
(i) a ∈
/ {0, b, (α + 1)b},
(iii) a = (α + 1)b 6= 0, (so that a 6= b,)
is satisfied, then hypothesis (H) is satisfied, so that the conclusion of the proposition holds.
(i) Suppose a = b. If a = 0, then (1, α, α + 1) is a solution of (e1 ). If a 6= 0, then (a, y, y) with
y∈
/ {0, a} is a solution of (e1 ). Thus, in the two cases, (e1 ) admits solutions (y1 , y2 , y3 ) ∈ F 3 such
that y1 y2 y3 6= 0 and y1 6= y2 . Hypothesis (H) is satisfied with u = 1.
(ii) Suppose a ∈
/ {0, b, (α + 1)b}. Then a + α(a + b) 6= 0. Let u = α, y3 = α(a + b). Choose
y1 ∈ F − {0, a + α(a + b)} and y2 = y1 + a + α(a + b). Then, y1 6= y2 and y1 y2 y3 6= 0, so that (H)
is satisfied.
(iii) Suppose a = (α+1)b 6= 0. Let u = (α+1), y3 = b. Choose y1 ∈ F −{0, αb} and y2 = y1 +αb.
Then, y1 6= y2 , y1 y2 y3 6= 0 and y1 + y2 + y3 = (α + 1)b = a, y1 + y2 + uy3 = αb + (α + 1)b = b, so
that (H) is satisfied.
(III) We examine the remaining case, that is the case a = 0, b 6= 0. Lemma 2.4 gives the
existence of (z1 , z2 , z3 ) ∈ F 3 , a solution of (S1 (b, c2 /b, d/b)) such that
2
b z1 +
bz12
b2 z12 + (α + 1)b2 z22 + αb2 z32 = c,
+ (α + 1)b2 z2 + bz22 + αb2 z3 + bz32 = d.
Let
u1 = b, u2 = (α + 1)b, u3 = αb, y1 = 1, y2 = α, y3 = α + 1.
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Then, (u1 , u2 , u3 , y1 , y2 , y3 , z1 , z2 , z3 ) is a solution of (D(b)) such that u1 u2 u3 y1 y2 y3 6= 0. Lemma 2.6. Let (a, b) ∈ F 2 . Then the system of equations
u1 + u2 + u3 = a,
(S2 (a, b))
x1 + x2 + x3 = b,
admits solutions (u1 , u2 , u3 , x1 , x2 , x3 ) ∈ F 6 satisfying the conditions
(2.1)
u1 u2 u3 6= 0,

(2.2)
1
det  u1
u1 x21
1
u2
u2 x22

1
u3  6= 0.
u3 x23 .
Proof. If (u1 , u2 , u3 , x1 , x2 , x3 ) ∈ F 6 is a solution of (S2 (0, 1)) satisfying conditions (2.1) and
(2.2), then for b ∈ F, b 6= 0, (u1 , u2 , u3 , bx1 , bx2 , bx3 ) is a solution of (S2 (0, b)) satisfying conditions (2.1) and (2.2). If (u1 , u2 , u3 , x1 , x2 , x3 ) ∈ F 6 is a solution of (S2 (1, 0)) satisfying conditions (2.1) and (2.2), then for a ∈ F, a 6= 0, (au1 , au2 , au3 , x1 , x2 , x3 ) is a solution of (S2 (a, 0))
satisfying conditions (2.1) and (2.2). If (u1 , u2 , u3 , x1 , x2 , x3 ) ∈ F 6 is solution of (S2 (1, 1)) satisfying conditions (2.1) and (2.2), then for a, b ∈ F, ab 6= 0, (au1 , au2 , au3 , bx1 , bx2 , bx3 ) is a
solution of (S2 (a, b)) satisfying conditions (2.1) and (2.2). It is sufficient to examine the cases
(a, b) = (0, 0), (a, b) = (0, 1), (a, b) = (1, 0), (a, b) = (1, 1). Observe that
(1, α, α2 , α, 1, α2 ) is a solution of (S2 (0, 0)) satisfying conditions (2.1) and (2.2);
(1, α, α2 , 0, 0, 1) is a solution of (S2 (0, 1)) satisfying conditions (2.1) and (2.2);
(1, α, α, 1, α, α2 ) is a solution of (S2 (1, 0)) satisfying conditions (2.1) and (2.2);
(1, α, α, 0, 0, 1) is a solution of (S2 (1, 1)) satisfying conditions (2.1) and (2.2).
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Lemma 2.7. Let (u1 , u2 , u3 , x1 , x2 , x3 ) ∈ F 6 be such that
u1 u2 u3 6= 0,
(2.1)
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(2.2)
1
det  u1
u1 x21
1
u2
u2 x22

1
u3  6= 0.
u3 x23 .
Then, for every (c, d) ∈ F 2 , there exists (y1 , y2 , y3 ) ∈ F 3 such that
y1 + y2 + y3 = c,
(S3 (c, d))
u21 y12 + u1 x21 y1 + . . . + u23 y32 + u3 x23 y3 = d.
Proof. Let N denote the number of (y1 , y2 , y3 ) ∈ F 3 solutions of (S3 (c, d)). With the notations
used in the proof of Lemma 2.4, we have
N=
X
(y1 ,y2 ,y3 )∈F 3
1X
Ψ(t(c + y1 + y2 + y3 ))
4
t∈F
1X
Ψ(u(d + u21 y12 + u22 y22 + u23 y32 )).
×
4
u∈F
Thus,
X
16N =
Ψ(ct + du)
(t,u)∈F 2
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Θi (t, u),
i=1
where
Θi (t, u) =
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3
Y
X
Ψ(ty + u(u2i y 2 + ui x2i y)).
y∈F
From [2, Proposition 2.3], Θi (t, u) ∈ {0, 4} and Θi (t, u) = 4 if and only if uu2i = (t + uui x2i )2 .
Thus,
X
N =4
Ψ(ct + du),
(t,u)∈E
where E is the set of pairs (t, u) ∈ F 2 such that

2
2

 t + uu1 x1 = u u1 ,
2
t + uu2 x2 = u2 u2 ,

 t + uu x2 = u2 u .
3 3
3
Observe that (0, 0) ∈ E. Moreover, if (t, 0) ∈ E, then t = 0. Suppose that (t, u) ∈ E with u 6= 0.
Then,
t = u(u1 x21 + uu1 ) = u(u2 x22 + uu2 ) = u(u3 x23 + uu3 ),
so that
u1 x21 + uu1 = u2 x22 + uu2 ,
u1 x21 + uu1 = u3 x23 + uu3 .
u1 x21 + u2 x22 = u(u1 + u2 ),
u1 x21 + u3 x23 = u(u1 + u3 ),
Thus,
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so that
(u1 x21 + u2 x22 )(u1 + u3 ) + (u1 x21 + u3 x23 )(u1 + u2 ),
in contradiction with condition (2.2).
Proposition 2.8. Let b = (b1 , b2 , . . . , b7 ) ∈ F 7 . Then the system of equations

3
P



ui = b1 ,


i=1



3
P



x = b2 ,

 i=1 i



3
P


(yi + u2i x2i ) = b3 ,



i=1

P
3
(E(b))
(zi + ui x3i ) = b4 ,

i=1


3

P



(u2i yi2 + ui x2i yi ) = b5 ,



i=1


3
P



(ui x2i zi + ui xi yi2 + u2i x2i ) = b6 ,



i=1


3

P


 (u2i zi2 + ui yi3 + u2i xi yi + ui x3i ) = b7
i=1
admits solutions (u1 , u2 , u3 , x1 , x2 , x3 , y1 , y2 , y3 , z1 , z2 , z3 ) ∈ F 12 with u1 u2 u3 6= 0.
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Proof. Lemma 2.6 gives the existence of (u1 , u2 , u3 , x1 , x2 , x3 ) ∈ F 6 , a solution of S2 (b1 , b2 )
satisfying (2.1) and (2.2). Lemma 2.7 gives the existence of (y1 , y2 , y3 ) ∈ F 3 , a solution of S3 (b3 +
3
P
u2i x2i , b5 ). Condition (2.2) insures the existence of (z1 , z2 , z3 ) ∈ F 3 such that
i=1








z1 + z2 + z3 = b4 +
3
P
i=1
3
P
ui x3i ,
(u2i x3i + u2i yi3 + ui x2i yi2 ,
u1 z1 + u2 z2 + u3 z3 = b27 +

i=1


3

P


 u1 x21 z1 + u2 x22 z2 + u3 x23 z3 = b6 +
(ui xi yi2 + u2i x2i ).
i=1
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Lemma 2.9. Let (a, b, c) ∈ F × × F 2 be such that ab + c 6= 0. Then the system


 u + v = a,
x + y = b,
(S4 (a, b, c))

 ux + vy = c
admits a solution (u, x, v, y) ∈ F 4 such that uv 6= 0 and u2 x + v 2 y 6= 0.
Proof. Let u ∈ F − {0, a} and v = u + a. Then uv(u + v) 6= 0, so that with x = (bu + c + ab)/a
and y = (bu + c)/a, (u, x, v, y) is a solution of (S4 (a, b, c)). Suppose that u2 x + v 2 y = 0. Then,
u2 b + uab + ac = 0. If b = 0, then c = 0, in contradiction with ab + c 6= 0. Thus b 6= 0, so that
c
2
2
u2 + au + ac
b = 0 and ab ∈ {0, 1}. Thus, c = 0. We have u x + v y = 0 and ux + vy = 0. Since
b 6= 0, we have (x, y) 6= (0, 0). If x = 0, then vy = 0, so that y = 0, a contradiction. Similarly,
y = 0 is impossible. Thus, xy =
6 0. Therefore u = u2 x/ux = v 2 y/vy = v in contradiction with
u + v 6= 0. Hence, (u, x, v, y) is a solution of (S4 (a, b, c)) such that uv 6= 0 and u2 x + v 2 y 6= 0. JJ J
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Proposition 2.10. Let b = (b1 , b2 , . . . , b8 ) ∈ F 8 . Then the system of equations

3

 P vi = b1 ,



 i=1


3

Pu =b ,

i
2


 i=1


3

 P (x + v 2 u2 ) = b ,

i
3

i i

 i=1


3

P

 (yi + vi u3i ) = b4 ,
i=1
(F(b))
3
P



(vi2 x2i + vi u2i xi + ui + zi ) = b5 ,


i=1



3
P



(vi u2i yi + vi ui x2i + vi2 u2i ) = b6 ,



i=1


3
P


(vi u3i + vi x3i + vi2 yi2 + vi u2i zi + vi2 ui xi ) = b7 ,



i=1


3
P


 (vi2 ui yi + vi ui yi2 + vi x2i yi ) = b8
i=1
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admits solutions
(v1 , v2 , v3 , u1 , u2 , u3 , x1 , x2 , x3 , y1 , y2 , y3 , z1 , z2 , z3 ) ∈ F 15
satisfying the condition v1 v2 v3 6= 0.
Proof. Let v1 = 1, v2 ∈ F − {0, 1, b1 + 1} and let v3 be defined by
v1 + v2 + v3 = b1 .
Then we have
v1 v2 v3 6= 0,
v1 6= v2 .
Let u1 ∈ F − {0, (v1 v2 )2 }, u3 = 0 and let u2 be defined by
u1 + u2 + u3 = b2 .
Then we have
v1 u21 6= v3 u23 .
(†)
(I) Suppose that v1 u1 + v2 u2 = 0. Let y1 ∈ F , y2 ∈ F − {u2 y1 /u1 } and let y3 be defined by
y1 + y2 + y3 = b4 +
3
X
vi u3i .
i=1
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Then we have
v1 v2 (u1 y2 + u2 y1 ) 6= 0,
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so that

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1

det  v1 u1
v1 y1
1
v2 u2
v2 y2

1

v3 u3  =
6 0.
v3 y 3
(II) Suppose that v1 u1 + v2 u2 6= 0. Let y1 ∈ F . Since u1 6= (v1 v2 )2 , we have 1 + v1 v2 u1 6= 0. Let
y2 ∈ F be such that
(1 + v1 v2 u1 )y2 + (1 + v1 v2 u2 )y1 6= v3 (v1 u1 + v2 u2 )(b4 +
3
X
i=1
and let y3 be defined by
y1 + y2 + y3 = b4 +
3
X
vi u3i .
i=1
Then we have
v1 v2 (u1 y2 + u2 y1 ) + v3 y3 (v1 u1 + v2 u2 ) 6= 0,
so that

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1

det  v1 u1
v1 y 1
1
v2 u2
v2 y 2

1

v3 u 3  =
6 0.
v3 y3
In both cases we get the existence of (y1 , y2 , y3 ) ∈ F 3 satisfying
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
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1
 2 2
det  v1 u1
v12 y12
1
2 2
v2 u2
v22 y22

1

v32 u23  =
6 0
v32 y32
vi u3i ),
from which we deduce the existence of (x1 , x2 , x3 ) ∈ F 3 such that
x1 + x2 + x3 = b3 +
3
P
i=1
vi2 u2i ,
v12 u21 x1 + v22 u22 x2 + v32 u23 x3 = (b1 + b2 + b6 )2
3
P
i=1
v12 y12 x1 + v22 y22 x2 + v32 y32 x3 = b28
3
P
vi u2i yi ,
(vi u2i yi2 + vi2 u2i yi ).
i=1
From (†),
det
1
v1 u21
1
v2 u22
6= 0.
Then there exists (z1 , z3 ) ∈ F 2 such that
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z1 + z3 = b2 + b5 +
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3
P
(vi2 x2i + vi u2i xi ),
i=1
v1 u21 z1
+
v3 u23 z3
= b7 +
3
P
i=1
(vi u3i + vi x3i + vi2 yi2 + vi2 ui xi ).
Let z2 = 0. Then, (v1 , v2 , v3 , u1 , u2 , u3 , x1 , x2 , x3 , y1 , y2 , y3 , z1 , z2 , z3 ) is a solution of (F(b)) satisfying v1 v2 v3 6= 0.
Proposition 2.11. Let b = (b1 , b2 , . . . , b9 ) ∈ F 9 . Then the system of equations
(G(b))
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
































































8
P
ui = b1 ,
i=1
8
P
x i = b2 ,
i=1
8
P
(yi + u2i x2i ) = b3 ,
i=1
8
P
(zi + ui x3i ) = b4 ,
i=1
8
P
(u2i yi2 + ui x2i yi ) = b5 ,
i=1
8
P
(ui x2i zi + ui xi yi2 + u2i x2i ) = b6 ,
i=1
8
P
(u2i zi2 + ui yi3 + u2i xi yi + ui x3i ) = b7 ,
i=1
8
P
(u2i yi zi + ui x2i zi + ui yi zi2 ) = b8 ,
i=1
8
P
(u2i xi zi + ui xi zi2 + ui yi2 zi ) = b9
i=1
Close
admits solutions (u1 , . . . , u8 , x1 , . . . , x8 , y1 , . . . , y8 , z1 , . . . , z8 ) ∈ F 32 such that u1 . . . u8 6= 0.
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Proof. Proposition 2.1 insures the existence of a solution (x1 , x2 , u1 , u2 ) of (A(b2 , b26 )) such that
u1 u2 x1 x2 6= 0. Thus, we have
x 1 + x 2 = b2 ,
u1 x21 z1
+
u1 x1 y12
+
u21 x21
+
u2 x22 z2
+
u2 x2 y22
+ u22 x22 = b6 .
Let y1 = y2 = z1 = z2 = 0. Proposition 2.5 insures the existence of a solution (u3 , u4 , u5 , y3 , y4 , y5 , z3 , z4 , z5
F 9 of (D((b3 + b6 , b25 , b29 , b8 ))) such that u3 u4 u5 y3 y4 y5 6= 0. Let x3 = x4 = x5 = 0. Then, we have
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 5
P


x i = b2 ,



i=1


5
P



(yi + u2i x2i ) = b3 ,



i=1


5

P


(u2i yi2 + ui x2i yi ) = b5 ,

i=1
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5
P



(ui x2i z1 + ui xi yi2 + u2i x2i ) = b6 ,


i=1



5
P



(u2i yi zi + ui x2i zi + ui yi zi2 ) = b8 ,



i=1


5

P


(u2i xi zi + ui xi zi2 + ui yi2 zi ) = b9 .
i=1
Let
β 1 = b1 +
5
X
ui ,
i=1
β 4 = b4 +
5
X
(zi + ui x3i ),
i=1
β 7 = b7 +
5
X
(u2i zi2 + ui yi3 + u2i xi yi + ui x3i ).
i=1
admits a solution (u6 , u7 , u8 , z6 , z7 , z8 ) ∈ F 6 such that
From Proposition 2.2,
u6 u7 u8 z6 z7 z8 6= 0. Let x6 = x7 = x8 = y6 = y7 = y8 = 0. Then, (u1 , . . . , u8 , x1 , . . . , x8 ,
y1 , . . . , y8 , z1 , . . . , z8 ) is a solution of (G(b)) such that u1 . . . u8 6= 0.
(B3 (β1 , β4 , β72 ))
3. Strict sums of degree less than 21 in F [T ]
The aim of this section is the proof of the three following theorems.
Theorem 3.1. Let A ∈ F [T ] with degree ≤ 7, say
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A=
7
X
ai T i .
i=0
Then, A is a strict sum of seventh powers if and only if its coefficients ai satisfy the conditions

 a1 = a4 ,
a2 = a5 ,
(3.1)

a3 = a6 .
Moreover, if A is a strict sum of seventh powers, then A is a strict sum of 5 seventh powers.
Theorem 3.2. Let A ∈ F [T ] with degree ≤ 14, say
A=
14
X
ai T i .
i=0
Then, A is a strict sum of seventh powers

 a1 + a4
a2 + a5
(3.2)

a3 + a6
if and only if its coefficients ai satisfy the conditions
+ a10 + a13 = 0,
+ a8 + a11 = 0,
+ a9 + a12 = 0.
Moreover, if A is a strict sum of seventh powers, then A is a sum of 11 seventh powers.
Theorem 3.3. Let A ∈ F [T ] be such that 15 ≤ deg A ≤ 21, say
A=
21
X
ai T i .
i=0
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Then, A is a strict sum of seventh powers if and only if its coefficients a1 , . . . , a21 satisfy the
condition
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(3.3)
a3 + a6 + a9 + a12 + a15 + a18 = 0.
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Moreover, if A satisfies condition (3.3), then A is a strict sum of 19 seventh powers.
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Theorem 3.1 is a consequence of the two following propositions.
Proposition 3.4. For (a, b, c) ∈ F 3 ,
cT 7 + (aT 2 + bT + c)(T 4 + T ) + a
= ((a + b + c)(T + 1))7 + ((α2 a + αb + c)(T + α))7
(3.4)
+ ((αa + α2 b + c)(T + α2 ))7 .
Proof. A verification.
Proposition 3.5.
(i) Let A ∈ F [T ] be such that deg A ≤ 6. If A is a strict sum of seventh powers, then its
coefficients satisfy (3.1).
(ii) Let
7
X
A=
ai T i
i=0
in the polynomial ring F [T ] be such that conditions (3.1) are satisfied. Then, A is a strict
sum of 5 seventh powers.
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Proof. Let A = a0 + a1 T + . . . + a6 T 6 ∈ F [T ]. Suppose that A is a strict sum of s seventh
powers. Then,
s
X
7
A=
(xi T + yi )
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i=1
with xi , yi ∈ F for i = 1, . . . , s. Thus,
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a1 = a4 , a2 = a5 , a3 = a6 .
3
Now let (a, b, c) ∈ F and let A = a7 T 7 + (T 4 + T )(aT 2 + bT + c) + a0 . From (3.4),
A + (a7 + c)T 7 + a0 + a = X17 + X27 + X37 ,
where X1 , X2 , X3 ∈ F [T ] have degree ≤ 1, so that
A = ((a7 + c)T )7 + (a0 + a)7 + X17 + X27 + X37 .
Corollary 3.6. We have S × (F, 7) 6= S(F, 7), so that g(4, 7) = ∞.
Proof. Conditions (3.1) are not satisfied by T , so that S × (F [T ], 7) 6= F [T ]. On the other hand,
from Paley’s theorem, [15], [6, Theorem 1.7], S(F [T ], 7) = F [T ].
Theorem 3.2 is a consequence of the following proposition.
Proposition 3.7. Let A ∈ F [T ] with degree ≤ 14, say A = a0 + a1 T + . . . + a14 T 14 .
(i) If A is a sum
A=
s
X
(Xi )7
i=1
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with Xi ∈ F [T ] of degree ≤ 2, then the cofficients a1 , . . . , a13 satisfy (3.2).
(ii) If (a1 , . . . , a13 ) ∈ F 13 satisfies (3.2), then A is a sum
7
A = X17 + . . . + X11
of 11 seventh powers of polynomials Xi with deg Xi ≤ 2.
Proof. (i) Suppose that A is a sum
A=
s
X
i=1
xi T 2 + yi T + zi
7
with xi , yi , zi ∈ F for i = 1, . . . , s. Then,
a1 + a4 + a10 + a13 =
s
X
yi (zi )3 +
i=1
+
s
X
s
X
(xi )2 (zi )2 + xi (yi )2 zi + yi (zi )3
i=1
s
X
(xi )2 (zi )2 + xi (yi )2 zi + (xi )3 (yi ) +
(xi )3 yi = 0.
i=1
i=1
The proof of the other identities is similar.
(ii) Conversely, suppose that (a1 , . . . , a13 ) ∈ F 13 satisfies (3.2). Proposition 2.3 insures the
existence of (x1 , x2 , y1 , y2 , z1 , z2 ) ∈ F 6 solution of (C(a11 , a13 , a9 )) such that x1 x2 y1 y2 6= 0. For
such a solution, we have

2
2
P
P


a13 =
yi =
x3i yi ,



i=1
i=1


2
2
P
P
a11 =
xi =
xi yi3 ,

i=1
i=1


2

P


 a9 =
(x2i yi2 + xi yi zi2 ).
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i=1
Let

2
P


a
=
a
+
(xi zi3 + x2i yi zi + xi yi3 ),

8


i=1


2
P
b = a12 +
(x3i zi + x2i yi2 ),

i=1


2

P


 c = a210 +
(xi zi + x2i yi zi2 + x3i yi2 ).
i=1
Proposition 2.2 gives the existence of a solution (x3 , x4 , x5 , z3 , z4 , z5 ) ∈ F 6 of (B3 (a, b, c)) such that
x3 x4 x5 z3 z4 z5 6= 0. For such a solution, we have

5
5
P
P


a=
xi =
xi zi3 ,



i=3
i=3


5
5
P
P
b=
zi =
x3i zi ,

i=3
i=3


5


 c2 = P xi zi .

i=3
Let
x6 = a14 +
5
X
xi ,
y3 = y4 = y5 = y6 = z6 = 0.
i=1
Thus, we have
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
6
P


a
=
(x3i zi + x2i yi2 ),

12


i=1


6
P
a10 =
(x2i zi2 + xi yi2 zi + x3i yi ),

i=1


6

P


 a8 =
(xi zi3 + x2i yi zi + xi yi3 ),
i=1
as well as

6
P


a
=
x3i yi ,

13


i=1


6
P
a11 =
xi yi3 ,

i=1


6

P


 a9 =
(x2i yi2 + xi yi zi2 ).
i=1
Let
(‡)
B =A+
6
X
(xi T 2 + yi T + zi )7 .
i=1
Then deg B ≤ 7. If
B=
7
X
bi T i ,
i=0
then,
b4 + b1 = a4 + a1 +
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6
X
(x2i zi2 + xi yi2 zi ),
i=1
6
X
b5 + b2 = a5 + a2 +
(xi zi3 + x2i yi zi ),
i=1
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b6 + b3 = a6 + a3 =
6
X
(x3i zi + xi yi zi2 ).
i=1
Condition (3.2) insures that
b4 + b1 = a13 + a10 +
6
X
(x2i zi2 + xi yi2 zi ) = 0,
i=1
6
X
b5 + b2 = a11 + a8 +
(xi zi3 + x2i yi zi ) = 0,
i=1
b6 + b3 = a12 + a9 =
6
X
(x3i zi + xi yi zi2 ) = 0,
i=1
so that (3.1) is satisfied by (b1 . . . , b6 ). Proposition 3.5 gives the existence of polynomials
X1 , . . . , X5 ∈ F [T ] of degree ≤ 1 such that
B=
5
X
Xi7 .
i=0
We conclude with (‡).
Theorem 3.3 is a consequence of the following proposition.
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Proposition 3.8. Let
A=
21
X
ai T i
i=0
be a polynomial in F [T ] with deg A ≤ 21. Then, A may be written as a sum
A=
s
X
i
(Xi )7
with Xi ∈ F [T ] of degree ≤ 3 if and only if its coefficients satisfy the condition
(3.3)
a3 + a6 + a9 + a12 + a15 + a18 = 0.
Moreover, if A satisfies condition (3.3), then A is a sum of 19 seventh powers of polynomials
Xi ∈ F [T ] of degree ≤ 3.
Proof. (I) Let A =
21
P
ai T i ∈ F [T ]. Suppose that A is a sum
i=0
A=
s
X
(ui T 3 + xi T 2 + yi T + zi )7
i=1
with ui , xi , yi , zi ∈ F for i = 1, . . . , s. Then we have
a3 + a6 + a9 + a12 + a15 + a18 = 0.
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(II) Let (a0 , a1 , . . . , a20 , a21 ) ∈ F 22 satisfying (3.3). We construct a representation of A as a
sum of seventh powers of polynomials of degree ≤ 3 in two steps.
(i) First step – From Proposition 2.11, there exists
(u1 , . . . , u8 , x1 , . . . , x8 , y1 , . . . , y8 , z1 , . . . , z8 ) ∈ F 32
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solution of (G(b)) with
b = (a21 , a20 , a19 , a18 , a17 + a20 , a16 , a15 , a1 + a4 + a10 + a13 + a16 + a19 ,
a2 + a5 + a8 + a11 + a17 + a20 ),
u1 . . . u8 6= 0. Therefore, we have
JJ J
 8
P


ui = a21 ,



i=1


8

P 3



ui xi = a20 ,


i=1



8
P


 (u3i yi + u2i x2i ) = a19 ,



i=1


8

P

 (u3i zi + ui x3i ) = a18 ,




 i=1
8
P
(u2i yi2 + ui x2i yi ) = a17 + a20 ,

i=1


8


 P (ui x2 zi + ui xi y 2 + u2 x2 ) = a16 ,


i
i
i i

 i=1


8

 P (u2 z 2 + u y 3 + u2 x y + u x3 ) = a ,

i i
i i
15

i i
i i i

 i=1


8

P 2

(u yi zi + ui x2i zi + ui yi zi2 ) = a1 + a4 + a10 + a13 + a16 + a19 ,


 i=1 i


8

P

 (u2i xi zi + ui xi zi2 + ui yi2 zi ) = a2 + a5 + a8 + a11 + a17 + a20 ,
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i=1
so that
a17 =
8
X
(u3i xi + u2i yi2 + ui x2i yi ).
i=1
(ii) Second step – Let
(?)
B =A+
8
X
(ui T 3 + xi T 2 + yi T + zi )7 .
i=1
Then deg B ≤ 14. If
B=
14
X
bi T i ,
i=0
then
b13 + b10 + b4 + b1 = a13 + a10 + a4 + a1 +
8
X
(ui xi yi2 + u2i yi zi + ui yi zi2 + u3i yi ),
i=1
b12 + b9 + b6 + b3 = a12 + a9 + a6 + a3 +
8
X
(u2i xi yi + u3i zi + u2i zi2 + ui yi3 ),
i=1
JJ J
8
X
b11 + b8 + b5 + b2 = a11 + a8 + a5 + a2 +
(u2i xi zi + ui xi zi2 + ui x2i yi + ui yi2 zi ,
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We have
a19 + a16 + a13 + a10 + a4 + a1 =
8
X
i=1
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+ u2i yi2 )
(u2i yi zi + ui x2i zi + ui yi zi2 ),
and
a19 + a16 =
8
X
(u3i yi + ui x2i zi + ui xi yi2 ),
i=1
so that
a13 + a10 + a4 + a1 =
8
X
(u2i yi zi + ui yi zi2 + ui xi yi2 + u3i yi ).
i=1
Thus,
b13 + b10 + b4 + b1 = 0.
Similarly, we prove that
b12 + b9 + b6 + b3 = b11 + b8 + b5 + b2 = 0.
Proposition 3.7 gives the existence of polynomials X1 , . . . , X11 ∈ F [T ] such that
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11
X
Xi7 ,
deg Xi ≤ 2.
i=1
We conclude with (?).
Remarks. Proposition 3.7 proves that T is not a sum of seventh powers of polynomials of degree
≤ 2. From Proposition 3.8 we deduce that every P ∈ F [T ] of degree ≤ 2 may be written as a
sum of 19 seventh powers of polynomials of degree ≤ 3, so that T is a sum of 19 seventh powers.
This gives another proof of the equality S(F [T ], k) = F [T ]. The following proposition gives a
representation of T as a sum of 12 seventh powers of polynomials of degree ≤ 3.
Proposition 3.9. We have
T = (T 3 + T 2 + 1)7 + (T 3 + T 2 + αT )7 + (T 3 + T 2 + (α + 1)T )7
+ (αT 3 + αT 2 + αT + α + 1)7 + ((α + 1)T 3 + (α + 1)T 2
+ (α + 1)T + α)7 + (T 2 + T + 1)7 + (T 2 + T )7 + (T 2 + α)7
+ (T 2 + α + 1)7 + (T + α)7 + (T + α + 1)7 + (T + 1)7 .
Proof. An easy verification.
4. The first descent
The process described in [1] or in [11] works when a representation of T as sum of k-th powers is
known. In the case when k = 7 and q = 4, this process leads to the following.
Theorem 4.1.
(i) Every polynomial P ∈ F [T ] with degree divisible by 7 and ≥ 18599 is a strict sum of 32
seventh powers.
(ii) Every polynomial P ∈ F [T ] with degree ≥ 18593 is a strict sum of 33 seventh powers.
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Proof. Let P ∈ F [T ] with 7(n − 1) < deg P ≤ 7n. Let
(
0 if deg P = 7n,
ε(P ) =
1 if deg P < 7n
and let
H = ε(P )T 7n + P.
Then, deg H = 7n. From [1, Lemma 5.2], there is a sequence H0 , H1 , . . . , Hi , . . . , of polynomials of F [T ] of degree 7n0 , 7n1 , . . . , 7ni , and a sequence X0 , X1 , . . . , Xi of polynomials of degree
n0 , n1 , . . . , ni , such that H = H0 and such that for each index i,
(4.1)
Hi = Xi7 + Hi+1 ,
(4.2)
6ni ≤ 7ni+1 < 6ni + 7.
Moreover, for each index i, there is a polynomial Yi ∈ F [T ] of degree ni such that
(4.3)
deg(Hi + Yi7 ) < 6ni .
We use (4.1) or (4.3) as long as the sequence (ni ) is decreasing. Let r, if it exists, be the least
index such that 3(6nr − 1) ≤ n. We use identity (1) r times, then we use identity (4.3) once. We
get
7
H = X07 + · · · + Xr−1
+ Yr7 + R,
with 3 deg R ≤ n. From Proposition 3.9, there exist R1 , . . . R12 ∈ F [T ] of degree ≤ 3 deg R such
that
7
R = R17 + . . . + R12
,
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so that
(4.4)
7
7
H = X07 + · · · + Xr−1
+ Yr7 + R17 + . . . + R12
with deg Xi = ni ≤ n0 = n, deg Yr = nr ≤ n0 = n, deg Rj ≤ 3 deg R ≤ n. Thus, (4.4) is a strict
sum of r + 13 seventh powers. From (4.2) we get that for i ≥ 1,
7i ni ≤ 6i n +
i−1
X
j=0
7j 6i−j .
Therefore, for any integer r ≥ 1, we have
6nr − 1 ≤ 6
For r ≥ 19, we have
6 r
7
<
1
18 .
r
r
6
6
n + 35 − 36
.
7
7
Suppose r = 19. If n ≥ 2657, then
19
19
6
6
n
6
n + 35 − 36
≤ .
7
7
3
5. Others descents
The second descent process is based on very simple identities.
Proposition 5.1. The following identity holds in the ring F [X, Y ],
(5.1)
X 4 Y 3 + XY 6 = X 7 + (X + Y )7 + (X + αY )7 + (X + (α + 1)Y )7 .
Proof. A simple verification.
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Proposition 5.2. For a non-negative integer i and X ∈ F [T ], let
Li (X) = X 4 T 3i + XT 6i .
(5.2)
Then, the map Li is F2 -linear and we have
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(5.3)
Li (X) = X 7 + (X + T i )7 + (X + αT i )7 + (X + (α + 1)T i )7 .
Close
(5.4)
T 7 Li (X) = Li+1 (T X).
Proof. Immediate.
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Corollary 5.3. Let n be a non-negative integer and let a ∈ F . Then, we have
(5.5)
aT 4n = L0 (aT n ) + aT n ,
(5.6)
aT 4n+3 = L1 (aT n ) + aT n+6 .
If n > 0, then
(5.7)
aT 4n+2 = L2 (aT n−1 ) + aT n+11 .
If n > 1, then
(5.8)
aT 4n+1 = L3 (aT n−2 ) + aT n+16 .
Proof. (5.5) and (5.6) are immediate. We get (5.7) and (5.8) noting that aT 4n+2 = aT 4(n−1)+6
and that aT 4n+1 = aT 4(n−2)+9 .
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Roughly speaking, the second descent process uses the following idea. Let X = xN T N +
xN −1 T N −1 + . . . + x1 T + x0 be a polynomial of F [T ]. Making use of (5.5)–(5.8), we replace a
monomial xk T k by the sum of an appropriate Li (T j ) and a monomial of lower degree. We begin
with xN T N and we follow decreasing degrees as long as the process gives monomials of lower
degree. For more details see [4, Proposition 5.4]. Mixing this process with the first descent process
leads to the following proposition.
Proposition 5.4. Let H ∈ F [T ] with degree 7n ≥ 112. Then, there exist X0 , X1 , X2 , X3 , Y0 , Y1 ,
Y2 , Y3 , Z ∈ F [T ] with deg Xi ≤ n, deg Yj ≤ n and deg Z ≤ 21 such that
(5.9)
H = X07 + X17 + X27 + X37 + L0 (Y0 ) + L1 (Y1 ) + L2 (Y2 ) + L3 (Y3 ) + Z.
Proof. See [4, Proposition 5.5].
We continue with other descent processes.
Proposition 5.5. Let n ≥ 3 be an integer and let A ∈ F [T ] of degree ≤ 7n. Then there exist
X1 , . . . , X4 ∈ F [T ] of degree ≤ n such that
!
3
X
7
Xi ≤ 7(n − 1),
deg A +
i=1
so that there exist X1 , . . . , X5 ∈ F [T ] of degree ≤ n such that
!
4
X
deg A +
Xi7 = 7(n − 1).
i=1
Proof. Let
A=
21
X
ai T i
i=0
be a polynomial of degree ≤ 21. Proposition 2.8 gives the existence of
(u1 , u2 , u3 , x1 , x2 , x3 , y1 , y2 , y3 , z1 , z2 , z3 ) ∈ F 16 ,
a solution of (E(a21 , a20 , a19 , a18 , a17 + a20 , a16 , a15 )) such that u1 u2 u3 6= 0.
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Therefore, we have

























3
P
ui = a21 ,
i=1
3
P
i=1
3
P
u3i xi = a20 ,
(u3i yi + u2i x2i ) = a19 ,
i=1
3
P
(u3i zi + ui x3i ) = a18 ,

i=1


3

P



(u3i xi + u2i yi2 + ui x2i yi ) = a17 ,



i=1


3
P



(ui x2i zi + ui xi yi2 + u2i x2i ) = a16 ,



i=1


3

P


(u2i zi2 + ui yi3 + u2i xi yi + ui x3i ) = a15 .

i=1
Let
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3
X
A+
(ui T 3 + xi T 2 + yi T + zi )7
!
,
i=1
so that
deg B ≤ 14.
This gives the first part of the proposition in the case when n = 3. We get the second part of the
proposition in the case n = 3 taking
0
if deg B = 14,
X4 =
T2
if deg B < 14.
Let n ≥ 3 be an integer and let A ∈ F [T ] of degree ≤ 7n. By euclidean division, there is a pair
(Q, R) ∈ F [T ] such that, respectively,
A = T 7(n−3) Q + R,
deg Q ≤ 21,
deg R < 7(n − 3).
There exist X1 , . . . , X3 ∈ F [T ] and X1 , . . . , X4 ∈ F [T ], of degree ≤ 3 such that
!
3
X
7
deg Q +
Xi ≤ 14,
i=1
and
deg Q +
4
X
!
Xi7
= 14,
i=1
respectively. Therefore,
deg T 7(n−3) (Q +
3
X
!
Xi7 )
≤ 7(n − 1),
i=1
and
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deg T
7(n−3)
(Q +
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4
X
!
Xi7 )
respectively, so that,
deg R + T
7(n−3)
Q+
3
X
i=1
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= 7(n − 1),
i=1
!
Xi7
≤ 7(n − 1),
and
deg R + T
7(n−3)
Q+
4
X
!
Xi7
= 7(n − 1),
i=1
respectively.
6. End of the proof
Proposition 6.1. Let
A=
28
X
ai T i
i=0
be a polynomial in F [T ] with deg A ≤ 28. Then A is a sum
(6.1)
A=
3
X
i=1
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Xi7
+
19
X
Yi7 ,
i=1
where X1 , . . . , X3 , Y1 , . . . , Y19 are polynomials of F [T ] such that deg Xi ≤ 4 and deg Yi ≤ 3.
Proof. Set
(6.2)
σ = a27 + a24 + a18 + a15 + a12 + a9 + a6 + a3 .
Proposition 2.10 gives the existence of
(v1 , . . . , v3 , u1 , . . . , u3 , x1 , . . . , x3 , y1 , . . . , y3 , z1 , . . . , z3 ) ∈ F 15 ,
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a solution of (F(a28 , a27 , a26 , a25 , a24 , a23 , a22 , σ) such that v1 . . . v4 6= 0. For such a solution, we
have

3

 a28 = P vi ,



i=1



3
3
P
P



a27 =
ui =
vi3 ui ,



i=1
i=1


3
3
P
P


2 2

a
=
(x
+
v
u
)
=
(vi3 xi + vi2 u2i ),
26
i

i
i


i=1
i=1


3
3

P
P


(yi + vi u3i ) =
(vi3 yi + vi u3i ),
 a25 =
i=1
i=1
(6.3)
3
P



(vi2 x2i + vi u2i xi + ui + zi )
 a24 =

i=1



3
P



=
(vi2 x2i + vi u2i xi + vi3 ui + vi3 zi ),



i=1


3

P


(vi u2i yi + vi ui x2i + vi2 u2i ),
 a23 =


i=1


3

P


(vi u3i + vi x3i + vi2 yi2 + vi u2i zi + vi2 ui xi ),
 a22 =
i=1
and
(6.4)
σ=
3
X
(vi2 ui yi + vi ui yi2 + vi x2i yi ).
i=1
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For i = 1, 2, 3, let
Xi = vi T 4 + ui T 3 + xi T 2 + yi T + zi
and let
B =A+
3
X
Xi7 .
i=1
Identities (6.3) show that deg B ≤ 21. Set
B=
21
X
bi T i .
i=1
From (6.2), (6.3) and (6.4),
b18 + b15 + b12 + b9 + b6 + b3 = 0,
so that from Theorem 3.3, there exist polynomials Y1 , . . . , Y19 ∈ F [T ] with degree ≤ 3 such that
B=
19
X
Yi7 .
i=1
Corollary 6.2. Let A ∈ F [T ] be such that 21 < deg A ≤ 28. Then A is a strict sum of 22
seventh powers.
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Theorem 6.3.
(i) Every polynomial P ∈ F [T ] whose degree ≥ 441 is divisible by 7 is a strict sum of 32 seventh
powers.
(ii) Every polynomial P ∈ F [T ] with degree ≥ 435 is a strict sum of 33 seventh powers.
(iii) Every polynomial P ∈ F [T ] such that deg P ≥ 112 and deg P is divisible by 7 is a strict
sum of 42 seventh powers.
(iv) Every polynomial P ∈ F [T ] with degree ≥ 106 is a strict sum of 43 seventh powers.
Proof. As for the proof of Theorem 4.1, it is sufficient to prove (i) and (iii). Let H ∈ F [T ] of
degree 7n with n ≥ 16. From (5.9) and (5.3), we get that there exists Z ∈ F [T ] with deg Z ≤ 21
such that H + Z is sum of 20 seventh powers of polynomials of degree ≤ n. From Proposition 3.9,
there exist Z1 , . . . , Z12 with deg Zi ≤ 63 such that
Z=
12
X
Zi7 .
i=1
If n ≥ 63, then H is a strict sum of 32 seventh powers. This proves (i).
From Proposition 6.1, there exist V1 , . . . , V22 ∈ F [T ] with deg Vi ≤ 4 < n such that
Z=
22
X
Vi7 ,
i=1
so that H is a strict sum of 42 seventh powers. This proves (iii).
Corollary 6.4. We have
G(4, 7) = G× (4, 7) ≤ 33.
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We end the study of the set S × (F, T ) dealing with polynomials P such that 29 ≤ deg P ≤ 105.
Proposition 6.5. Let A ∈ F [T ].
(i) If 29 ≤ deg A ≤ 35, then A is a strict sum of 25 seventh powers.
(ii) If deg A = 42, then A is a strict sum of 26 seventh powers.
(iii) If 35 < deg A < 42, then A is a strict sum of 27 seventh powers.
(iv) If deg A = 7n with 7 ≤ n < 14, then A is a strict sum of n + 20 seventh powers.
(v) If 7n − 7 < deg A < 7n with 7 ≤ n < 14, then A is a strict sum of n + 21 seventh powers.
(vi) If deg A = 7n with 14 ≤ n < 21, then A is a strict sum of n + 19 seventh powers.
(vii) If 7n − 7 < deg A < 7n with 14 ≤ n < 21, then A is a strict sum of n + 20 seventh powers.
(viii) If deg A = 7n with 21 ≤ n < 28, then A is a strict sum of n + 18 seventh powers.
(ix) If 7n − 7 < deg A < 7n with 21 ≤ n < 28, then A is a strict sum of n + 19 seventh powers.
Proof. As observed before, it suffices to prove (i), (ii), (iv), (vi) and (viii).
1. Suppose that 29 ≤ deg A ≤ 35. From Proposition 5.5, there exist X1 , X2 , X3∈ F [T ] of degree
3
P
≤ 5 such that deg(A +
Xi7 ) ≤ 28. From Proposition 6.1, there exist Y1 , . . . , Y22 ∈ F [T ] of degree
i=1
≤ 4 such that
A+
3
X
Xi7 =
i=1
22
X
Yj7 .
j=1
2. Suppose that deg A = 42. From [1, Lemma 5.2-(i)], there is a polynomial X ∈ F [T ] of degree
6 such that deg(A + X 7 ) ≤ 35. From above, there exist Y1 , . . . , Y25 ∈ F [T ] of degree ≤ 5 such that
A + X7 =
25
X
Yj7 .
j=1
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3. We prove (iv), (vi) and (viii) by induction. Suppose that for n ≥ 7, every polynomial of
degree 7k with k < n is a strict sum of s(k) seventh powers. Let A ∈ F [T ] of degree 7n. From [1,
Lemma 5.2-(ii)], there is a polynomial X ∈ F [T ] of degree n such that deg(A + X 7 ) = 7m(n) with
m(n) defined by the condition 6n ≤ 7m(n) < 6n + 7. We have

if 7 ≤ n ≤ 13,
 n−1
n−2
if 14 ≤ n ≤ 20
m(n) =

n−3
if 21 ≤ n ≤ 27.
The induction hypothesis gives that A + X 7 is a strict sum of s(m(n)) seventh powers, so that A
is a strict sum of s(m(n)) + 1 seventh powers. We have s(6) = 26. Thus,

if 7 ≤ n ≤ 13,
 n + 20
n + 19
if 14 ≤ n ≤ 20,
s(n) =

n + 18
if 21 ≤ n ≤ 27.
Proposition 6.6. We have
S × (F [T ], 7) = A1 ∪ A2 ∪ A3 ∪ A∞ ,
where
(i) A1 is the set of polynomials A =
(ii) A2 is the set of polynomials A =
7
P
n=0
14
P
an T n ∈ F [T ] such that a1 = a4 , a2 = a5 , a3 = a6 ,
an T n ∈ F [T ] with 7 < deg A ≤ 14 such that
n=0
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
 a1 + a4 + a10 + a13 = 0,
a2 + a5 + a8 + a11 = 0,

a3 + a6 + a9 + a12 = 0,
(iii) A3 is the set of polynomials A =
21
P
an T n ∈ F [T ] with 14 < deg A ≤ 21 such that
n=0
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a3 + a6 + a9 + a12 + a15 + a18 = 0,
(iv) A∞ = {A ∈ F [T ] | deg A > 21}.
Proof. With Theorems 3.1, 3.2, 3.3, Corollary 6.2 and Theorem 6.3.
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Theorem 6.7. We have
g × (4, 7) ≤ 43.
Proof. From Theorems 3.1, 3.2, 3.3, every polynomial A ∈ S × (F [T ], 7) of degree ≤ 21 is a
strict sum of 19 seventh powers. From Corollary 6.2 and Proposition 6.5, every polynomial A ∈
S × (F [T ], 7) such that 21 < deg A ≤ 175 is a strict sum of 43 seventh powers. From Theorem 6.3,
every polynomial A ∈ S × (F [T ], 7) such that deg A ≥ 106 is a strict sum of 43 seventh powers. JJ J
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2.
, Sums of fourth powers of polynomials over a finite field of characteristic 3, Func. and Approx,
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, Sums of (2r + 1)-th powers in the polynomial ring F2m [T ], Port. Math. (N.S), 67(1) (2010), 13–56.
, Sums of seventh powers in the polynomial ring F2m [T ], Port. Math. (N.S), 68(3) (2011), 297–316.
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5. Car M. and Gallardo L. H., Sums of cubes of polynomials, Acta Arith. 112 (2004), 41–50.
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7. Ellison W. J., Waring’s problem, Amer. Math. Monthly 78(1) (1971), 10–36.
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Mireille CAR, Aix-Marseille Université Faculté des Sciences et Techniques, Case cour A Avenue Escadrille NormandieNiemen, F-13397 Marseille Cedex 20, France,
e-mail: mireille.car@univ-amu.fr
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